Ta có: \(y'=3x^2-4x+4=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)+\dfrac{8}{3}=3\left(x-\dfrac{2}{3}\right)^2+\dfrac{8}{3}\ge\dfrac{8}{3}\forall x\in R\)
⇒ y'min = 8/3 tại x0 = 2/3
⇒ y0 = -79/27
⇒ PTTT: \(y=\dfrac{8}{3}\left(x-\dfrac{2}{3}\right)-\dfrac{79}{27}\)